combinatorics - Prove that $frac{100!}{50!cdot2^{50}} in Bbb{Z}$

I'm trying to prove that :

$$\frac{100!}{50!\cdot2^{50}}$$

is an integer .

For the moment I did the following :

$$\frac{100!}{50!\cdot2^{50}} = \frac{51 \cdot 52 \cdots 99 \cdot 100}{2^{50}}$$

But it still doesn't quite work out .

Hints anyone ?

Thanks

Answer

$$\frac{(2n)!}{n! 2^{n}} = \frac{\prod\limits_{k=1}^{2n} k}{\prod\limits_{k=1}^{n} (2k)} = \prod_{k=1}^{n} (2k-1) \in \Bbb{Z}.$$